Question

In the expansion of ${(1+x+x^2+x^3)}^6,$ the coefficient of $x^{14}$ is

• A. 130
• B. 120
• C. 128
• D. 125
• E. 115

Answer 1

Answer: (B)

${(1+x+x^2+x^3)}^6$
$={(1+x)}^6{(1+x^2)}^6$
$={(}^6{C}_0{+}^6{C}_{1}x{+}^6{C}_2{x}^2{+}^6{C}_3{x}^3{+}^6{C}_4{x}^4$ ${+}^6{C}_5{x}^5{+}^6{C}_6{x}^6)$ $\times {(}^6{C}_0{+}^6{C}_1{x}^2{+}^6{C}_2{x}^4{+}^6{C}_3{x}^6{+}^6{C}_4{x}^8$ ${+}^6{C}_5{x}^{10}{+}^6{C}_6{x}^{12})$
$\therefore$ Coefficient of $x^{14}$ in ${(1+x+x^2+x^3)}^6$ ${=}^6{C}_2{.}^6{C}_6{+}^6{C}_4{.}^6{C}_5{+}^6{C}_6{.}^6{C}_4$
$=\frac{6!}{2!4!}.\frac{6!}{0!6!}+\frac{6!}{4!2!}.\frac{6!}{5!1!}+\frac{6!}{0!6!}.\frac{6!}{2!4!}$
$=15+90+15=120$